Задание 69

Question

Решите уравнения:
ewlineа) $\frac{1}{2 - x} + \frac{1}{4} = -\frac{4}{x(2 - x)}$;
ewlineб) $\frac{7}{x^2 + 3x - 9} + x^2 + 3x = 1$.

Gimli · Accepted Answer

a) $$\frac{1}{2 - x} + \frac{1}{4} = -\frac{4}{x (2 - x)}$$
$$\frac{1}{2 - x} + \frac{4}{x (2 - x)} = -\frac{1}{4}$$
$$\frac{x + 4}{x (2 - x)} = -\frac{1}{4}$$
$$4x + 16 = -2x + x^2$$
$$x^2 - 2x - 4x - 16 = 0$$
$$x^2 - 6x - 16 = 0$$
$$D = 6^2 + 16	imes 4 = 36 + 64 = 100 = 10^2$$
$$x_1 = \frac{6 + 10}{2} = \frac{16}{2} = 8;  x_2 = \frac{6 - 10}{2} = \frac{-4}{2} = -2$$	extbf{Ответ: $х = {-2; 8}$}б) $$\frac{7}{x^2 + 3x - 9} + x^2 + 3x = 1$$Cделаем замену $x^2 + 3x$ на у:$$\frac{7}{y - 9} + y = 1$$
$$\frac{7 + y^2 - 9y}{y - 9} = 1$$
$$y^2 - 9y + 7 = y - 9$$
$$y^2 - 10y + 16 = 0$$
$$D = 100 - 64 = 36 = 6^2$$
$$y_{1} = \frac{10 + 6}{2} = 8; y_{2} = \frac{10 - 6}{2} = 2$$A теперь, используя значения у, найдём значения х:$$x^2 + 3x = 8$$
$$x^2 + 3x - 8 = 0$$
$$D = 3^2 + 32 = 9 + 32 = 41$$
$$x_{1; 2} = \frac{-3\pm \sqrt{41}}{2}$$
$$x^2 + 3x = 2$$
$$x^2 + 3x - 2 = 0$$
$$D = 9 + 8 = 17$$
$$x_{3; 4} = \frac{-3\pm \sqrt{17}}{2}$$	extbf{Ответ: $x = {\frac{-3\pm\sqrt {41}}{2}; \frac{-3\pm \sqrt{17}}{2}}$
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Задание 69

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Задание 69

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